A tetrapeptide, " x " on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is :

Solution

Chemistry ma’am · 3 steps

A peptide has an N-end and a C-end, so every order of the four amino acids counts: that is 4!.

1. What is fixed. A tetrapeptide has four amino acid units. Equimolar Gly, Ala, Val and Leu on hydrolysis means each of the four appears exactly once, one in each position.
2. Order matters. A peptide runs from a free -NH₂ end (N-terminus) to a free -COOH end (C-terminus). So Gly-Ala-Val-Leu and Leu-Val-Ala-Gly are different molecules, and every arrangement counts separately.

   Sequences = 4! = 4 × 3 × 2 × 1 = 24

3. So. There are 4 choices for the first position, 3 for the second, 2 for the third and 1 for the last. That gives 24 tetrapeptides, option (b).

The idea to keep. For a peptide of n different amino acids, each used once, the number of sequences is n!. The N to C direction makes every order a different peptide.

Answer from the NTA final answer key. Solution by PrepWiser.